Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct segt {
  double l, r;
};
double sy, a, b;
int n;
segt s[200111];
double pref[200111];
const double eps = 1e-8;
bool outleft(double x, double y, int idx) {
  double xcoord = x - (y - sy) * (x - s[idx].l) / y;
  return a - xcoord >= -eps;
}
bool outright(double x, double y, int idx) {
  double xcoord = x - (x - s[idx].r) * (y - sy) / y;
  return xcoord - b >= -eps;
}
double solve(double x, double y) {
  int L = 0, R = n + 1;
  double ret = 0;
  int low = 1, high = n;
  while (low <= high) {
    int mid = (low + high) >> 1;
    if (outleft(x, y, mid)) {
      L = mid;
      low = mid + 1;
    } else {
      high = mid - 1;
    }
  }
  low = 1, high = n;
  while (low <= high) {
    int mid = (low + high) >> 1;
    if (outright(x, y, mid)) {
      R = mid;
      high = mid - 1;
    } else {
      low = mid + 1;
    }
  }
  if (L + 1 <= R - 1) {
    ret += (y - sy) / y * (pref[R - 1] - pref[L]);
  }
  double delta = 0;
  if (L != 0) {
    delta += max(0.0, min(b, x - (x - s[L].r) * (y - sy) / y) - a);
  }
  if (R != n + 1) {
    delta += max(0.0, b - max(x - (y - sy) * (x - s[R].l) / y, a));
  }
  ret += min(delta, b - a);
  return ret;
}
int main() {
  int i;
  scanf("%lf%lf%lf", &sy, &a, &b);
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) {
    scanf("%lf%lf", &s[i].l, &s[i].r);
    pref[i] = pref[i - 1] + s[i].r - s[i].l;
  }
  int q;
  scanf("%d", &q);
  while (q > 0) {
    double x, y;
    scanf("%lf%lf", &x, &y);
    printf("%.15lf\n", solve(x, y));
    q--;
  }
  return 0;
}